Using elementary operations, find the inverse of the following matrix:

Let A = [[1, 2, 3], [0, 1, 5], [0, 0, 1]]
For applying elementary row operation we write, A = I.A
[[1, 2, 3], [0, 1, 5], [0, 0, 1]] = [[1, 0, 0], [0, 1, 0], [0, 0, 1]] A
Applying R1 ↔ R2, we get
[[1, 2, 3], [0, 1, 5], [0, 0, 1]] = [[0, 1, 0], [1, 0, 0], [0, 0, 1]] A
Applying R2 → R2 + R1 and R3 → R3 − R1, we get
[[1, 2, 3], [0, 3, 5], [0, 0, 1]] = [[0, 1, 0], [1, 1, 0], [0, -2, 1]] A
Applying R1 → R1 − 2/3R2, we get
[[1, 0, 1/3], [0, 3, 5], [0, 0, 1]] = [[−2/3, 1/3, 0], [1, 1, 0], [0, −2, 1]] A
Applying R2 → 1/3R2, we get
[[1, 0, 1/3], [0, 1, 5/3], [0, 0, 1]] = [[-2/3, 1/3, 0], [1/3, 1/3, 0], [0, -2, 1]] A
Applying R3 → R3 + 5R2, we get
[[1, 0, 1/3], [0, 1, 5/3], [0, 0, 1]] = [[-2/3, 1/3, 0], [1/3, 1/3, 0], [5/3, -1, 1]] A
Applying R1 → R1 − 1/3R3 and R2 → R2 − 5/3R3
[[1, 0, 0], [0, 1, 0], [0, 0, 1]] = [[−1, 1, −1/3], [−8/3, 6/3, −5/3], [5/3, -1, 1]] A
Applying R3 → 3R3, we get
[[1, 0, 0], [0, 1, 0], [0, 0, 1]] = [[-1, 1, -1/3], [-8/3, 2, -5/3], [5, -3, 3]] A
Hence A⁻¹ = [[-1, 1, -1/3], [-8/3, 2, -5/3], [5, -3, 3]]